1
$\begingroup$

How to solve this differential equation $\dot{x}=|x|$?

$\endgroup$
  • $\begingroup$ @Rahul because initial condition is not given. $\endgroup$ – bellcircle Nov 19 '18 at 11:10
5
$\begingroup$

One observes that $x=0$ is the trivial solution of the given ODE.

Now find nontrivial solutions of the equation.

Using separation of variables, one gets $$\frac{\dot{x}}{|x|}=1.$$

Integrating both sides gives

\begin{align}\text{sgn}(x)\log|x|=t+C\quad\cdots\quad\text{(a)}\end{align} for some constant $C$, where $\text{sgn}(x)=\begin{cases}1,&x> 0\\-1,&x<0.\end{cases}$

Therefore, $x(t)=\begin{cases}C_1e^t,&x>0\\C_2e^{-t},&x<0.\end{cases}$

Now find the solution of explicit form.

From the equation, $\dot{x}$ is always nonnegative, so $x$ is a continuously differentiable and nondecreasing function of $t$. This shows that $C_1$ must be positive and $C_2$ must be negative.

Since $x$ is a continuous function of $t$ and a nontrivial solution of $x$ cannot take 0 by the above observation, it follows that $x(t)=C_1e^t,C_1>0$ for all $t$ or $x(t)=C_2e^{-t},C_2<0$ for all $t$.

$\endgroup$
  • $\begingroup$ You first divided both sides by $|x|$ to get the following results. But how can you guarantee that you are not dividing by $0$. In other words, if you cannot guarantee that once the solution starts from a non-zero value, it will not reach $0$, then you cannot divide it by $|x|$. $\endgroup$ – winston Nov 19 '18 at 12:30
  • $\begingroup$ @winston Good question. If $\dot{x}=0$, then $x=0$ at such point. Now it is obvious that there is no way to combine the distinct (local) solutions $C_1e^t,C_2e^{-t},0$ to make a continuous global solution because the first is always positive, the second always negative, the last always zero. Therefore the global solution must be one of the three. $\endgroup$ – bellcircle Nov 19 '18 at 12:47
3
$\begingroup$

bellcircle has provided a solution to the problem. However, I would like to express the solution in a more compact form and also provide an answer using a proposition-proof style.

$(Dx)(t) = |x(t)|$ (1)

Proposition. If $I$ is an open interval in $\mathbb{R}$ and $C \in \mathbb{R}$, then $x(t)=Ce^{\mathsf{sgn}(C) t}$ is a solution of the differential equation (1) on $I$. If $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $\mathbb{R}$, then there exists $C \in \mathbb{R}$ such that $x(t)=Ce^{\mathsf{sgn}(C) t}$ for all $t \in I$.

Suppose $I$ is an open interval in $\mathbb{R}$. Suppose $C \in \mathbb{R}$ and $x$ is such that $x(t)=Ce^{\mathsf{sgn}(C) t}$ for all $t \in I$. Then $(Dx)(t)=\mathsf{sgn}(C)Ce^{\mathsf{sgn}(C) t}$. Also, $|x(t)|=\mathsf{sgn}(C)Ce^{\mathsf{sgn}(C) t}$ for all $t \in I$. Therefore, indeed $x(t)=Ce^{\mathsf{sgn}(C) t}$ is a solution of (1) on $I$.

Suppose there is a solution $x_2(t)$ of (1) on $I$ such that there is no $C \in \mathbb{R}$ such that $x_2(t) = Ce^{\mathsf{sgn}(C) t}$ for all $t \in I$. Then take $T \in I$. Define $A=\frac{x_2(T)}{e^{\mathsf{sgn}(x_2(T)) T}}$. Then $x(t) = A e^{\mathsf{sgn}(A)t}$ is a solution of (1) on $I$ with $x(T)=x_2(T)$. The uniqueness of the solution of (1) on $I$ with $x(T)=x_2(T)$ follows from the Lipschitz continuity of $|\cdot|$ by the Picard–Lindelöf theorem (link) and the global uniqueness theorem (e.g. link). Therefore $x_2(t) = A e^{\mathsf{sgn}(A)t}$ for all $t \in I$ and, thus, by contradiction, if $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $\mathbb{R}$, then there exists $C \in \mathbb{R}$ such that $x(t)=Ce^{\mathsf{sgn}(C) t}$ for all $t \in I$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.